TITLE: Integral $\int_1^\infty\operatorname{arccot}\left(1+\frac{2\pi}{\operatorname{arcoth}x-\operatorname{arccsc}x}\right)\frac{\mathrm dx}{\sqrt{x^2-1}}$
QUESTION [79 upvotes]: Consider the following integral:
$$\mathcal{I}=\int_1^\infty\operatorname{arccot}\left(1+\frac{2\,\pi}{\operatorname{arcoth}x\,-\,\operatorname{arccsc}x}\right)\frac{\mathrm dx}{\sqrt{x^2-1}}\,,$$
where $\operatorname{arccsc}$ is the inverse cosecant, $\operatorname{arccot}$ is the inverse cotangent and $\operatorname{arcoth}x$ is the inverse hyperbolic cotangent.
Approximate numerical integration suggests a possible closed form:
$$\mathcal{I}\stackrel?=\frac{\pi\,\ln\pi}4-\frac{3\,\pi\,\ln2}8.$$
I was not able to rigorously establish the equality, but the value is correct up to at least $900$ decimal digits.

Is it the correct exact value of the integral $\,\mathcal{I}$?

REPLY [7 votes]: I shall transform the given integral into the previously-proven identity:
$$\int_0^1\frac{\operatorname{arccot}\left(1+\frac\pi{\operatorname{arctanh}x-\arctan x}\right)}x\,dx=\frac{\pi\ln\pi}4-\frac{3\pi\ln2}8$$
As was shown by @Chen Wang, the given integral is equal to
$$\begin{align}\mathcal{I}&=4\pi \int_{0}^{1} \frac{z^2 \ln (z)}{(z^4-1)(\pi^2+(2\operatorname{arctanh}(z)+2\operatorname{arccot}(z))^2)} \, dz \\
&=\Im \int_{0}^{1} \frac{4z^2\ln (z)}{(z^4-1)(2\operatorname{arctanh} (z)+2\operatorname{arccot} (z)-i\pi)} \, dz \\
&\stackrel{\text{IBP}}{=}\int_{0}^{1} \frac{1}{z} \Im \left(\ln(2\operatorname{arctanh} (z)+2\operatorname{arccot}(z)-i\pi)+\frac{\pi i}{4}\right) \, dz \end{align}$$
Since $\displaystyle\small\Im \left(\ln(2\operatorname{arctanh} (z)+2\operatorname{arccot}(z)-i\pi)+\frac{\pi i}{4}\right)=\Im \left(\ln\left(1+\frac{1+i}{\pi} (\operatorname{arctanh} (z)-\operatorname{arctan}(z))\right)\right)$, this implies that
$$\mathcal{I}=\int_{0}^{1} \frac{1}{z} \Im \left(\ln\left(1+\frac{1+i}{\pi} (\operatorname{arctanh} (z)-\operatorname{arctan}(z))\right)\right)\, dz$$
which is identically the same as the previously-proven identity, and so the conjectured closed-form is true.
$\square$